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Here's the problem:

Prove from the definition of a limit that $$ \lim\limits_{x \to -1} \frac{x+2}{x+3}=\frac{1}{2} $$

Ok, I just started doing these, so bear with me. I've set up these guys:

$$ \lvert x+1 \rvert \lt \delta \text{ and } \left\lvert \frac{x+2}{x+3} -\frac{1}{2} \right\rvert \lt \epsilon $$.

Now, I'm used to doing simpler examples in which I just do some algebra to the "epsilon" side to get it to look like the "delta" side, but doesn't really work here. So, basically, I'm stuck. I've tried all of the expanding and term-collecting that I can do, and nothing's popping out for me. Any help here would be appreciated. Thanks.

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For $\epsilon>0$ given, take $\delta=\min(\epsilon,1)$. Then, if $|x-(-1)|=|x+1|<\delta$, we have $$|x+3|=|2+(x+1)|\geq 2-|x+1|>2-\delta\geq1$$ so $$\left|\frac{x+2}{x+3}-\frac{1}{2}\right|=\left|\frac{x+1}{2(x+3)}\right|=\frac{|x+1|}{2|x+3|}<\frac{\epsilon}{2}<\epsilon$$

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  • $\begingroup$ By the way, in general we have $|a+b|\geq|a|-|b|$. This follows from the triangle inequality by $|a|=|(a+b)-b|\leq|a+b|+|-b|=|a+b|+|b|$. $\endgroup$ – Spenser Mar 20 '15 at 21:56
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Hint: Rewrite $$\frac{x+2}{x+3}-\frac{1}{2}=\frac{x+1}{2(x+3)}$$

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